Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}8x+4y &= 8 \\ -3x-4y &= 1\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-3x = 4y+1$ Divide both sides by $-3$ to isolate $x$ $x = {-\dfrac{4}{3}y - \dfrac{1}{3}}$ Substitute this expression for $x$ in the first equation. $8({-\dfrac{4}{3}y - \dfrac{1}{3}}) + 4y = 8$ $-\dfrac{32}{3}y - \dfrac{8}{3} + 4y = 8$ Simplify by combining terms, then solve for $y$ $-\dfrac{20}{3}y - \dfrac{8}{3} = 8$ $-\dfrac{20}{3}y = \dfrac{32}{3}$ $y = -\dfrac{8}{5}$ Substitute $-\dfrac{8}{5}$ for $y$ in the top equation. $8x+4( -\dfrac{8}{5}) = 8$ $8x-\dfrac{32}{5} = 8$ $8x = \dfrac{72}{5}$ $x = \dfrac{9}{5}$ The solution is $\enspace x = \dfrac{9}{5}, \enspace y = -\dfrac{8}{5}$.